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The distance between (2, 1) and (n, 4) is 5 units. Find all possible values of n.

User Tamera
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1 Answer

5 votes

Answer:

6 and -2

Explanation:

To find the possible values of n, we can use the distance formula between two points in a coordinate plane.

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

In this case, we are given the points (2, 1) and (n, 4), and the distance is 5 units. Plugging these values into the distance formula, we get:

5 = √[(n - 2)² + (4 - 1)²]

Simplifying the equation, we have:

25 = (n - 2)² + 9

25 = n² - 4n + 4 + 9

25 = n² - 4n + 13

Rearranging the equation, we have:

n² - 4n - 12 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring the equation, we have:

(n - 6)(n + 2) = 0

Setting each factor equal to zero, we get:

n - 6 = 0 or n + 2 = 0

Solving for n in each case, we find:

n = 6 or n = -2

Therefore, the possible values of n are 6 and -2.

User Alexis
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